Question: The equation of a circle $C$ is $x^2+y^2-12x-14y+69 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Answer: To find the equation in standard form, complete the square. $(x^2-12x) + (y^2-14y) = -69$ $(x^2-12x+36) + (y^2-14y+49) = -69 + 36 + 49$ $(x-6)^{2} + (y-7)^{2} = 16 = 4^2$ Thus, $(h, k) = (6, 7)$ and $r = 4$.